You are Here: Home >< Physics

# Mechanics-How do you do this ? Watch

1. Posted from TSR Mobile
Attached Images

2. Look at the triangle with an angle in it. The opposite side to angle has length=3.5m.
S=3.5m U=? V=0ms a-9.81 t=?
V^2=U^2+2aS
U=√(-2 X (-9.81) X3.5)
U= 8.28673639 vertical component of initial velocity

You can work out time of flight for reaching the top of the wall. You know horizontal distance between ramp and wall. Since the horizontal component of velocity is constant, you can use speed=distance/time to work out the horizontal component of velocity. Then you'd have both vertical and horizontal components of velocity, allowing you to determine initial speed of the stunt rider by using Pythagoras' Theorem.
3. I can't get one of the answer options but:
For projectile motion you separate the horizontal and vertical components so that:

Vertical: Horizontal:
S: 3.5 S: 10
U: ? U: ?
V: 0 V: X
a: -9.81 a: 0
t: ? t: ?

The using the SUVAT equations: V squared = U squared + 2as the vertical component of u can be found as 8.29 m per s.
Then the time taken to reach the half way point can be found using V = U + at as 0.85 s.
Then as t for the vertical component is equal to that of t for the horizontal component: S = Ut + 0.5at squared
As a is equal to zero: S= ut, which is 10 = u x 0.85 s. Therefore U is equal to 11.76 m per s.

Then using a squared + be squared = c squared the initial velocity can be found. 11.76 squared + 8.29 squared = 207.02.
The square root of that is 14.4 m per s. But that is not offered on the answer options.
4. a=(v-u)/t
t=(v-u)/a
t= 8.28673639/9.81
t= 0.84472338 time of flight to reach top of wall

Speed=distance/time
speed=10/ 0.84472338
Speed= 11.8381949 initial velocity horizontal component

From earlier 8.28673639ms= vertical component of initial velocity

Velocity off ramp=√( 8.28673639^2 + 11.8381949^2 )
=
14.4503584207360145 ms

I'm afraid I can't get the right answer as well then
5. Isn't the answer then 25ms since we assumed no air resistance above an 25 is closest to 14.45... and is also above 14.45...?
6. I would go with that one (25.1) as the question doesn't tell you to ignore air resistance.
7. (Original post by krisshP)
Look at the triangle with an angle in it. The opposite side to angle has length=3.5m.
S=3.5m U=? V=0ms a-9.81 t=?
V^2=U^2+2aS
U=√(-2 X (-9.81) X3.5)
U= 8.28673639 vertical component of initial velocity
This is correct.

Edit:

Next bit incorrect - apologies.

Theta is not the ramp angle.

My Original post

So if this is the vertical component then 8.28673639 = V sin θ ---- (1)
where V is the actual velocity of projection and theta the angle shown to the horizontal.

You can work out θ from the triangle with 3.5 and 10m
tan θ = 3.5/10 ------ (2)

If you know theta you can find V in (1)

It's a lesson in not doing questions in a hurry when it's late, and not assuming the given answer is correct.
Apologies.
Stonebridge
8. Thanks
9. (Original post by Stonebridge)
This is correct.

So if this is the vertical component then 8.28673639 = V sin θ ---- (1)
where V is the actual velocity of projection and theta the angle shown to the horizontal.

You can work out θ from the triangle with 3.5 and 10m
tan θ = 3.5/10 ------ (2)

If you know theta you can find V in (1)
I get
25.0846851

But the thing I can't understand is why using Pythagoras theorem can't work like above where I said

Speed=distance/time speed=10/ 0.84472338
Speed= 11.8381949 initial velocity horizontal component

From earlier 8.28673639ms= vertical component of initial velocity

Velocity off ramp=√( 8.28673639^2 + 11.8381949^2 ) = 14.4503584207360145 ms

The horizontal component of velocity is constant and we have initial velocity vertical component, so surely this method should work, why not?
Thanks
10. (Original post by krisshP)
I get
25.0846851

But the thing I can't understand is why using Pythagoras theorem can't work like above where I said

Speed=distance/time speed=10/ 0.84472338
Speed= 11.8381949 initial velocity horizontal component

From earlier 8.28673639ms= vertical component of initial velocity

Velocity off ramp=√( 8.28673639^2 + 11.8381949^2 ) = 14.4503584207360145 ms

The horizontal component of velocity is constant and we have initial velocity vertical component, so surely this method should work, why not?
Thanks
Oops
I've looked at this again this morning. It was a bit late last night.
I was wrong.
Theta is of course, NOT, the angle of the ramp. The ramp angle must be greater than that.
Your method is correct because you need to work out the time taken to reach the 3.5m height.
Using this time and the horizontal distance gives the horizontal component, and thus the two components give the actual velocity of projection.
The problem is, that doesn't give any of the allowed answers.
I suspect the person setting this has made the same mistake I did.
This is what happens when you
a) rush things when it's too late
b) assume the given answers must include the correct one
I've edited my earlier post to point this out.
11. (Original post by Stonebridge)
Oops
I've looked at this again this morning. It was a bit late last night.
I was wrong.
Theta is of course, NOT, the angle of the ramp. The ramp angle must be greater than that.
Your method is correct because you need to work out the time taken to reach the 3.5m height.
Using this time and the horizontal distance gives the horizontal component, and thus the two components give the actual velocity of projection.
The problem is, that doesn't give any of the allowed answers.
I suspect the person setting this has made the same mistake I did.
This is what happens when you
a) rush things when it's too late
b) assume the given answers must include the correct one
I've edited my earlier post to point this out.
So then is the correct answer 14.4503584207360145 ms? If it is, it does make some sense to select the answer in the choices which is greater then 14.4503584207360145ms and is closest to it since no air resistance is assumed and also the stunt rider would want to generally have a speed greater than minimum speed needed as a kind of safety net if that makes sense. So I guess it is reasonable to say that option 25.1ms should be chosen right?

But then the answer of 25.0846851ms with the incorrect method rounds to 25.1ms which is an option

Nice question OP
12. (Original post by krisshP)
So then is the correct answer 14.4503584207360145 ms? If it is, it does make some sense to select the answer in the choices which is greater then 14.4503584207360145ms and is closest to it since no air resistance is assumed and also the stunt rider would want to generally have a speed greater than minimum speed needed as a kind of safety net if that makes sense. So I guess it is reasonable to say that option 25.1ms should be chosen right?

But then the answer of 25.0846851ms with the incorrect method rounds to 25.1ms which is an option

Nice question OP
None of the options is correct.
I would bet that the person who prepared or worked out the answer to this question has made the same mistake that I did.

The main problem with it is that the wall has a definite thickness.
In order to clear the wall, you need to rise higher than the 4m height of the side.
So the value of 3.5m is actually not quite the correct number for the suvat formula to calculate v or t
Having said that there is no way of knowing what value to use without knowing the thickness of the wall. If you did have this info the question would be a lot more complicated and I'm certain this was not the intention of the question.
Labelling the angle theta was very misleading, too. But it's not my excuse for making the mistake.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 17, 2013
Today on TSR

### What is the latest you've left an assignment

And actually passed?

### Simply having a wonderful Christmas time...

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.